Inverse functions of one-to-one functions

[echkae]

For A I got f^-1(x) = cubic root (x) + 3
For B I got f^-1(x) = square root (x-6) + 1
For C I got f^-1(x) = cos^-1 [(x+3)/4]
For D I got f^-1(x) = ln (x-2)/ln 3

I'm not confident in my answers though.

 

[ksdhart2]

For the benefit of other users, I've transcribed your image:

Find the inverse functions of each one-to-one function:

A. $\displaystyle f(x) = (x-3)^3$
B. $\displaystyle g(x) = x^2 - 2x + 5$, where $\displaystyle x \ge 1$
C. $\displaystyle h(x) = 2cos(2x) - 3$, where $\displaystyle 0 \le x \le \pi$
D. $\displaystyle f(x) = 3^{x-3} + 2$

With that out of the way... in the future, if you're ever unsure of an answer you can always check it yourself. In this case, we'll use the definition of inverse functions, that $\displaystyle f(f^{-1}(x)) = f^{-1}(f(x)) = x$. I'll work through checking one answer and leave you to check the rest.

$\displaystyle f(f^{-1}(x)) = (\left[ \sqrt[3]{x} + 3 \right]-3)^3$

$\displaystyle = \left[ \sqrt[3]{x} + 3 \right]^3 - 9\left[ \sqrt[3]{x} + 3 \right]^2 + 27\left[ \sqrt[3]{x} + 3 \right] - 27$

$\displaystyle = \left[ 9x^{2/3} + 27x^{1/3} + x + 27 \right] - \left[ 9x^{2/3} + 54x^{1/3} + 81 \right] + \left[ 27x^{1/3} + 81 \right] - 27$

$\displaystyle = 9x^{2/3} + 27x^{1/3} + x + 27 - 9x^{2/3} - 54x^{1/3} - 81 + 27x^{1/3} + 81 - 27$

$\displaystyle = x$

Okay, that's good. Let's check the other way around just to be sure though:

$\displaystyle f^{-1}(f(x)) = \sqrt[3]{(x-3)^3} + 3$

$\displaystyle = (x-3) + 3$

$\displaystyle = x$

This also checks out, so therefore we know you're answer is correct.

 

[lookagain]

I'll work through checking one answer and leave you to check the rest.

$\displaystyle f(f^{-1}(x)) = (\left[ \sqrt[3]{x} + 3 \right]-3)^3$

$\displaystyle = \left[ \sqrt[3]{x} + 3 \right]^3 - 9\left[ \sqrt[3]{x} + 3 \right]^2 + 27\left[ \sqrt[3]{x} + 3 \right] - 27$

$\displaystyle = \left[ 9x^{2/3} + 27x^{1/3} + x + 27 \right] - \left[ 9x^{2/3} + 54x^{1/3} + 81 \right] + \left[ 27x^{1/3} + 81 \right] - 27$

$\displaystyle = 9x^{2/3} + 27x^{1/3} + x + 27 - 9x^{2/3} - 54x^{1/3} - 81 + 27x^{1/3} + 81 - 27$

$\displaystyle = x$

Okay, that's good.

No, ksdhart2, that is not how the first one is to be done. It's on the same level of complexity as the work for $\displaystyle f^{-1}(f(x))$ that you showed below it.



$\displaystyle f(f^{-1}(x)) = (\left( \sqrt[3]{x} + 3 \right)-3)^3$

$\displaystyle = ( \sqrt[3]{x} + 3 - 3)^3$

$\displaystyle = ( \sqrt[3]{x})^3$

$\displaystyle = x$

 

[JeffM]

No, ksdhart2, that is not how the first one is to be done. It's on the same level of complexity as the work for $\displaystyle f^{-1}(f(x))$ that you showed below it.



$\displaystyle f(f^{-1}(x)) = (\left( \sqrt[3]{x} + 3 \right)-3)^3$

$\displaystyle = ( \sqrt[3]{x} + 3 - 3)^3$

$\displaystyle = ( \sqrt[3]{x})^3$

$\displaystyle = x$

Well, it is A logically valid way for it to be done, but somewhat slower and more error prone than your way.

$\displaystyle \left ( \left \{ \sqrt[3]{x} + 3 \right \} - 3 \right )^3 = \left ( -\ 3 + \left \{x^{1/3} + 3 \right \} \right )^3 =$

$\displaystyle 1(-\ 3)^3(x^{1/3}+ 3)^0 + 3(-\ 3)^2(x^{1/3} + 3)^1 + 3(-\ 3)^1(x^{1/3} + 3)^2 + 1(-\ 3)^0(x^{1/3} + 3)^3=$

$\displaystyle -\ 27 + 27(x^{1/3} + 3) - 9(x^{1/3} + 3)^2 + (x^{1/3} + 3)^3 =$

$\displaystyle 54 + 27x^{1/3} - 9x^{2/3} - 54x^{1/3} - 81 + x^{3/3} + 9x^{2/3} + 27x^{1/3} + 27 =$

$\displaystyle (54 - 81 + 27) + x^{1/3}(27 - 54 + 27) + x^{2/3}(9 - 9) + x^{3/3} = x^{3/3} = x.$

Nothing logically incorrect. But admittedly burdensome.

 

[mmm4444bot]

For B I got f^-1(x) = square root (x-6) + 1
For C I got f^-1(x) = cos^-1 [(x+3)/4]
For D I got f^-1(x) = ln (x-2)/ln 3

My answers differ, but not by much.

Double-check your steps. There may be some simple mistakes.

For B, did you use Completing the Square? If so, you might have subtracted (b/2)^2 from each side, instead of adding it.

For C, were you thinking that 1/2*arccos([x+3]/2) = arccos([x+3]/4)? We can't multiply like that.

For D, it seems like you might have dropped a constant term, somewhere.

If you cannot find any errors in your steps, please post your work. :cool: